views:

253

answers:

2

xml:

<entry>
  <link rel="http://schemas.google.com/g/2005#feed" type="application/atom+xml" href="http://picasaweb.google.com/data/feed/api/user/xy/albumid/531885671007533108" /> 
  <link rel="alternate" type="text/html" href="http://picasaweb.google.com/xy/Cooking" /> 
  <link rel="self" type="application/atom+xml" href="http://picasaweb.google.com/data/entry/api/user/xy/albumid/531885671007533108" /> 
</entry>

Here's what I've tried:

foreach($xml->entry as $feed) {
 $album_url = $feed->xpath("./link[@rel='alternate']/@href");
 echo $album_url;
}

I've tried all kinds of permutations, too but no luck.

Expected result would be http://picasaweb.google.com/xy/Cooking

The result I get is "". Can someone explain what I'm doing wrong?

Can someone please help me out? I've been at this for hours...

+1  A: 

You were close:

./link[@rel='alternate']/@href

Should be the correct XPath to get those values.

Jacob
Hey Jacob, thanks for the answer... I tried to use your entry as above, but it still is returning no value.
Jay
Josh Davis's answer is correct; xpath returns multiple values. You could just be getting back an array.
Jacob
+1  A: 

xpath() returns an array, you have to select the first element of that array, at index 0. Attention: if there's no match, it may return an empty array. Therefore, you should add an if (isset($xpath[0])) clause, just in case.

foreach ($xml->entry as $entry)
{
    $xpath = $entry->xpath('./link[@rel="alternate"]/@href');

    if (isset($xpath[0]))
    {
        echo $xpath[0], "\n";
    }
}
Josh Davis